Define $L(x) = \sum_{n\leq x} \lambda(n)$, where $\lambda$ denotes the Liouville function. If $c$ is the supremum of the real parts of the zeros of the Riemann zeta function and $\lfloor x\rfloor$ denotes the integer part of $x$, is it true that $$\Bigl\lvert\int_{1}^{x} L(y)\lfloor x/y\rfloor\frac{\mathrm{d}y}{y} \Bigr\rvert > x^{c-\epsilon}$$ for arbitrarily large $x$ and any $\epsilon>0$?
2
-
2$\begingroup$ $\int_{1}^{x} L(y)\lfloor x/y\rfloor\frac{\mathrm{d}y}{y} = \int_1^x \frac{\lfloor \sqrt{t}\rfloor}{2t}dt$ $\endgroup$– reunsCommented Feb 10, 2020 at 1:54
-
$\begingroup$ It does not seem to be true. If you try $x=10^4$ and $\epsilon=10^{-5}$, either directly in your inequality, ,or in the inequality obtained from the comment above, Mathematica will tell you this is not true. You have to be careful with your inequality because the integral fails to converge adequately. $\endgroup$– EGMECommented Jun 26, 2022 at 17:59
Add a comment
|